Question

The largest value of $2x^3 - 3x^2 - 12x + 5$ for $-2 \leq x \leq 4$ occurs at $x$ is equal to :

• A. -4
• B. 4
• C. 1
• D. 0

Answer 1

Answer: (B)

4

Answer 2

$f(x)=2 x^{3}-3 x^{2}-12 x+5$ and $-2 \leq x \leq 4$ To find maxima, differentiate $f ( x )$ & put it equal to $o$ $f(x)=6 x^{2}-6 x-12=0$ $\Rightarrow x^{2}-x-2=0$ $\Rightarrow x^{2}-2 x+x-2=0$ $\Rightarrow x(x-2)+1(x-2)=0$ $(x-2)(x+1)=0$ $x=2,-1$ $f''(x)=12 x-6$ $f(2)=18 > 0$ $\therefore$ At $x=2,$ value of $f(x)$ is minimum $f(-1)=-18 < 0$ $\therefore x =-1$ can be point of maxima $\therefore$ We check value of $f(x)$ at $x=-2,2,-1,4$ $f (-2)=-16-12+24+5=1$ $f(2)=16-12-24+5=-15$ $f (-1)=-2-3+12+5=12$ $f(4)=128-48-48+5=37$ $\therefore$ At $x =4, f ( x )$ is maximum.